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A deterministic interpretation with a definite particle configuration guided by the wavefunction. The latter obeys the Time-dependent Schrödinger equation, while the particle follows the guidance equation. A Born rule distribution of initial positions is preserved by quantum equilibrium equivariance. The guidance law restricts admissible initial velocities, even when trajectories are rewritten as a second-order force equation.

Writing the wavefunction in modulus and dimensionless quantum phase gives a probability continuity equation and a Hamilton-Jacobi equation with a quantum potential:The phase equations hold locally away from wavefunction nodes. They follow by equating real and imaginary parts of the Time-dependent Schrödinger equation.

For unit mass and $\hslash =1$, this prepoint kernel tends to the identity and generates the Time-dependent Schrödinger equation. The square-root branch follows the damped Fresnel integral. Its normalized moments are $1$, $0$, and $i\Delta t$ at orders zero, one and two; these fix both the normalization and the kinetic sign.

Expand the wavefunction in the complete orthonormal energy eigenstates. The Time-dependent Schrödinger equation makes each coefficient satisfy $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$, soThe series is interpreted in the Hilbert space norm. Its norm is conserved because the phase factors have modulus one and the energy eigenstates are orthonormal. For initial data in the domain of the Hamiltonian operator, this is a strong solution of the equation; for any square-summable coefficients it gives the corresponding unitary time evolution.

For the proposed Gaussian wave packet, differentiation givesInserting these in the Time-dependent Schrödinger equation with Hamiltonian $(p^2-x^2)/2$, its right-hand side becomesEquating the constant and quadratic coefficients with $i\hslash {\partial }_t\Psi$ yieldsConversely these two equations make the substitution an identity, so they are sufficient for the Gaussian evolution in an inverted harmonic oscillator.

The Time-dependent Schrödinger equation is $i\hslash {\partial }_t\Psi =H\Psi$. The normalized eigenstates of the Hermitian Hamiltonian $H_3$ with distinct eigenvalues are orthogonal, so they form an orthonormal basis of this two-state system. Substitute the expansion of $\Psi$ and compare the two coefficients:Hence $a_0(t)=a_0(0)e^{it/\hslash }$ and $a_1(t)=a_1(0)e^{-it/\hslash }$. For the normalized test state $\varphi =\alpha {\chi }_0+\beta {\chi }_1$, the Born rule givesThe complex conjugates are essential; relative phases can affect this probability even though the energy-state populations remain constant under $H_3$.

For a real potential, the Time-dependent Schrödinger equation and its complex conjugate giveThe potential terms cancel. The expression on the right is a total derivative, givingFor the plane wave, ${\psi }_t=-i\omega \psi$ and ${\psi }_{xx}=-k^2\psi$. With zero potential, substitution givesIt is a free-particle momentum eigenstate with momentum $\hslash k$, energy $\hslash \omega$ and uniform probability flux. On the whole real line it is a generalized state, not a normalized square-integrable wavefunction; wave packets or box normalization give the corresponding physical interpretation.

For the wavefunction of a particle of mass $m$, the probability density and probability current areFor a real potential $V$, the Time-dependent Schrödinger equation and its complex conjugate readMultiplying by ${\Psi }^{\ast }$ and $\Psi$ respectively yieldsThe potential terms cancel because $V$ is real. This is the continuity equation, expressing local conservation of probability.

For the given superposition the common time phase cancels in ${\Psi }^{\ast }{\Psi }_x$, andThe expression in the second parentheses is purely imaginary, so the second term is real and contributes nothing to the probability current. ThusIt is the incoming probability current minus the reflected probability current.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

Use the dimensionless quantum phase $S$, so that $\psi =R{e}^{iS}$ with $R\ge 0$. Work locally away from wavefunction nodes, where $R$ and $S$ are differentiable. Differentiating the wavefunction for substitution in the Time-dependent Schrödinger equation gives:Cancel $e^{iS}$ in the time-dependent equation and equate real and imaginary parts. The resulting Madelung equations areThe second equation is a Hamilton-Jacobi equation for the action $\hslash S$, with an additional quantum potential. To see the meaning of the first, multiply it by $2R$ and set $\rho =R^2$. It becomes the probability continuity equationThus the probability density is transported by the velocity field $\hslash \nabla S/m$. The Madelung equations are a local rewriting of the linear wave equation; their apparent nonlinearity comes from expressing a complex wavefunction in modulus and phase variables.

Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent isEach momentum integration is an ordinary Gaussian integral:Therefore the measure written in the question gives exactlyThe resulting configuration integral is $\int \mathcal{D}{q}_{\mathrm{raw},N}{e}^{-S_{E,N}}$. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses $d{p}_i/(2\pi )$ rather than $d{p}_i$. With that normalization the configuration measure isThis differs from the raw measure by $(2\pi {)}^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\dots ,q_N$ are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit $d{q}_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.

We now address the unheaded real-time continuation. Set $m=\hslash =1$. The normalized short-time Schrödinger kernel gives the final-slice recurrenceThe square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's $d{q}^{\prime }/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i{)}^{-1/2}$ fixes the identity limit and is included at every step.

Put $\eta =q^{\prime }-q$. The normalized oscillatory Gaussian has momentsTaylor-expand the smooth wavefunction and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order $(\Delta t{)}^2$. ThusSubtract the initial value, divide by $\Delta t$ and take the limit:This proves the Time-dependent Schrödinger equation by the Schrödinger equation from a short-time path integral argument. Nonuniform partitions give the same limit when their largest time step tends to zero.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrödinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrödinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

The Time-dependent Schrödinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.

The one-dimensional Time-dependent Schrödinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

For the one-dimensional quantum harmonic oscillator,This is the Time-dependent Schrödinger equation with harmonic potential.

Expand the normalized initial wavefunction in the orthonormal energy basis:The Time-dependent Schrödinger equation then gives the unitary time evolution

The Time-dependent Schrödinger equation is $i\hslash {\partial }_t\psi =\hat{H}\psi$. Differentiate $⟨\hat{A}⟩=⟨\psi |\hat{A}|\psi ⟩$, use the equation and its adjoint, and include explicit time dependence of the operator. This gives Ehrenfest theorem

Insert the stated expansion into the Time-dependent Schrödinger equationWhen the derivative acts on each free phase, its energy factor is cancelled by the action of $H_0$ on the corresponding eigenstate. The remaining terms obeyDivision by $i\hslash$ gives